If α = 2 6^ + 126^ 3 , then which of the following quadratic...

We are given $\alpha = \dfrac{2 \cos 6^\circ + \cos 126^\circ}{\sqrt{3}}$.

Split the term $2 \cos 6^\circ$ to apply the sum-to-product formula:

$\alpha = \dfrac{\cos 6^\circ + \cos 6^\circ + \cos 126^\circ}{\sqrt{3}}$

Using the formula $\cos C + \cos D = 2 \cos\left(\dfrac{C+D}{2}\right) \cos\left(\dfrac{C-D}{2}\right)$:

$\cos 6^\circ + \cos 126^\circ = 2 \cos 66^\circ \cos 60^\circ = 2 \cos 66^\circ \left(\dfrac{1}{2}\right) = \cos 66^\circ$

Substitute this back into the expression for $\alpha$:

$\alpha = \dfrac{\cos 6^\circ + \cos 66^\circ}{\sqrt{3}}$

Apply the sum-to-product formula again:

$\cos 6^\circ + \cos 66^\circ = 2 \cos 36^\circ \cos 30^\circ = 2 \cos 36^\circ \left(\dfrac{\sqrt{3}}{2}\right) = \sqrt{3} \cos 36^\circ$

Thus, $\alpha = \dfrac{\sqrt{3} \cos 36^\circ}{\sqrt{3}} = \cos 36^\circ$.

We know that $\cos 36^\circ = \dfrac{\sqrt{5} + 1}{4}$.

Therefore, $\alpha = \dfrac{\sqrt{5} + 1}{4}$

$\Rightarrow 4\alpha - 1 = \sqrt{5}$

Squaring both sides:

$(4\alpha - 1)^2 = 5$

$16\alpha^2 - 8\alpha + 1 = 5$

$16\alpha^2 - 8\alpha - 4 = 0$

$4\alpha^2 - 2\alpha - 1 = 0$

So, $\alpha$ is a root of the equation $4x^2 - 2x - 1 = 0$.

Answer: $4x^2 - 2x - 1 = 0$